Let $\def\A{\mathbb A}\def\F{\mathbb F}\F_3$ be the Galois field with three elements and let $\A^d=\A^d(\F_3)$ be the affine space of dimension $d$ over $\mathbb F_3$ —the subject is combinatorics and not algebraic geometry so this is essentially just the set $\F_3^d$ with the incidence structure whose blocks are the affine subspaces.

Suppose $p\in\A^d$ is point and $X\subseteq\A^d$ is a set not containing $p$ which intersects every line through $p$ in exactly one point.

Is it true that $X$ contains an affine subspace of dimension $d-1$?

If this is true, then in fact $X$ is a disjoint union $P_0\cup P_1\cup\cdots\cup P_{d-1}$ where each $P_i$ is an affine subspace of $\A^d$ of dimension $i$, and this decomposition is unique.

I think that I have a (horrid) proof of this but I know that there are people who are immensely more fluent in this sort of thing than I...

Notice that if one projectivizes $\mathbb A^d$ (with origin set at $p$) then one knows that the resulting projective space $\mathbb P^{d-1}(\mathbb F_3)$ does have a «cellular decomposition» and the question is asking if one can find one such decomposition which lifts to a decomposition of $X$ —the converse is obvious.
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Mariano Suárez-Alvarez♦May 10 '13 at 5:23

Isn't it possible to just compare the number of such disjoint unions vs. the number of such sets?
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François BrunaultMay 10 '13 at 8:22

If I'm not mistaken, the number of disjoint unions $P_0 \cup \cdots \cup P_{d-1}$ satisfying your condition is $(3^d-1) (3^{d-1}-1) \cdots (3-1)$, and the number of possible $X$'s is $2{\^}((3^d-1)/2)$ which is larger for $d \geq 3$.
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François BrunaultMay 10 '13 at 8:37

For an explicit counterexample when $d≥3$, you could start with an $X$ that is of the form $P_0\cup\dots\cup P_{d−1}$, and just replace one of the points of the $P_{d−1}$ with the other point on the same line through $p$.
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Jeremy RickardMay 10 '13 at 11:15